Step by step
The question
Question: a sample of 49 has mean 98.2, and the SD of all the numbers is 0.9: at the 5% level, is the true mean different from 98.6?
Alpha, 0.05, is the level: a sample rarer than 5% counts as too rare.
1 Write the hypotheses
H₀ is the "nothing new" idea: it has the equal sign.
Here: the mean is 98.6.
Hₐ is what we want to show.
Here: the mean is different from 98.6.
Different from means too big or too small: both ends of the bell count.
Two tails.
2 The test number
We act as if H zero is true: the mean is 98.6.
How far is our sample mean from it?
First the SD of the sample means: σ over the square root of n.
The square root of 49 is 7.
0.9/7 is 0.1286.
z is the sample mean minus 98.6, over 0.1286.
98.2 minus 98.6 is −0.4.
Divided by 0.1286: about −3.11.
3 The p-value
The p-value is the chance of getting a sample this far out, if H zero is true.
On the bell of H zero: 98.6 in the middle, our sample at z = −3.11.
Two tails: we shade both ends, as far out as our sample or more.
The z-table gives the area to the left of z.
Row −3.1, column .01.
For z = −3.11: the row −3.1 and the column .01 meet at 0.0009.
Careful!
0.0009 is only one end.
Two tails: we double it.
One end is 0.0009.
Two ends: 2 times that is 0.0018.
4 Decide
The p-value, 0.0018, is less than alpha, 0.05.
Our sample would be rare if H zero were true.
So we reject H zero.
5 Say it in words
In words: at the 5% level, there is enough evidence that the mean is different from 98.6.
6 Check
Let's check the other way, with the cut-off.
0.05 split into two ends gives the cut-offs plus and minus 1.96.
z is −3.11: past the cut-off.
The same decision.
So it is right!
Let's look back.
First: write the hypotheses.
Then: the test number.
Then: the p-value.
Then: decide.
Then: say it in words.
Last: check.